Grace is going to invest \( \$ 650 \) and leave it in an account for 15 years. Assuming the interest is compounded continuously, what interest rate, to the nearest tenth of a percent, would be required in order for Grace to end up with \( \$ 1,490 \) ?

To find the required interest rate, we can use the formula for continuous compounding, which is given by: \[ A = Pe^{rt} \] Where: - \( A \) is the amount of money accumulated after time \( t \), - \( P \) is the principal amount (the initial amount of money), - \( r \) is the annual interest rate (in decimal), - \( t \) is the time the money is invested for (in years), - \( e \) is the base of the natural logarithm, approximately equal to 2.71828. For Grace's situation: - \( A = 1490 \), - \( P = 650 \), - \( t = 15 \). We need to solve for \( r \): \[ 1490 = 650e^{15r} \] First, divide both sides by \( 650 \): \[ \frac{1490}{650} = e^{15r} \] Calculating the left side: \[ 2.29230769231 = e^{15r} \] Next, take the natural logarithm of both sides: \[ \ln(2.29230769231) = 15r \] Now, calculate \( \ln(2.29230769231) \): \[ \ln(2.29230769231) \approx 0.832909 \] Now we can solve for \( r \): \[ r = \frac{0.832909}{15} \approx 0.05552726 \] To express \( r \) as a percentage, multiply by 100: \[ r \approx 5.55\% \] Rounding to the nearest tenth of a percent, the required interest rate is approximately **5.6%**.